Consider a normed vector space \((X, \Vert \cdot \Vert)\). If \(X\) is finite-dimensional, then a subset \(Y \subset X\) is compact if and only if it is closed and bounded. In particular a closed ball \(B_r[a] = \{x \in X \, ; \, \Vert x – a \Vert \le r\}\) is always compact if \(X\) is finite-dimensional.

The first one is based on open covers. For \(n \ge 1\), we denote by \(f_n\) the piecewise linear map defined by \[
\begin{cases}
f_n(0)=f_n(\frac{1}{2^n}-\frac{1}{2^{n+2}})=0 \\
f_n(\frac{1}{2^n})=1 \\
f_n(\frac{1}{2^n}+\frac{1}{2^{n+2}})=f_n(1)=0
\end{cases}\] All the \(f_n\) belong to \(B_1[0]\). Moreover for \(1 \le n < m\) we have \(\frac{1}{2^n}+\frac{1}{2^{n+2}} < \frac{1}{2^m}-\frac{1}{2^{m+2}}\). Hence the supports of the \(f_n\) are disjoint and \(\Vert f_n – f_m \Vert = 1\).

Now consider the open cover \(\mathcal U=\{B_{\frac{1}{2}}(x) \, ; \, x \in B_1[0]\}\). For \(x \in B_1[0]\}\) and \(u,v \in B_{\frac{1}{2}}(x)\), \(\Vert u -v \Vert < 1\). Therefore, each \(B_{\frac{1}{2}}(x)\) contains at most one \(f_n\) and a finite subcover of \(\mathcal U\) will contain only a finite number of \(f_n\) proving that \(A\) is not compact.

Second proof based on convergent subsequence. As \(A\) is a metric space, it is enough to prove that \(A\) is not sequentially compact. Consider the sequence of functions \(g_n : x \mapsto x^n\). The sequence is bounded as for all \(n \in \mathbb N\), \(\Vert g_n \Vert = 1\). If \((g_n)\) would have a convergent subsequence, the subsequence would converge pointwise to the function equal to \(0\) on \([0,1)\) and to \(1\) at \(1\). As this function is not continuous, \((g_n)\) cannot have a subsequence converging to a map \(g \in A\).

Riesz’s theorem

The non-compactness of \(A=C([0,1],\mathbb R)\) is not so strange. Based on Riesz’s lemma one can show that the unit ball of an infinite-dimensional normed space \(X\) is never compact. This is sometimes known as the Riesz’s theorem.

The non-compactness of \(A=C([0,1],\mathbb R)\) is just standard for infinite-dimensional normed vector spaces!

Theorem (Brouwer): Every continuous function from a convex compact subset \(K\) of a Euclidean space to \(K\) itself has a fixed point.

We here describe an example of a homeomorphism of the unit ball of a Hilbert space having no fixed point. Let \(E\) be a separable Hilbert space with \((e_n)_{n \in \mathbb{Z}}\) as a Hilbert basis. \(B\) and \(S\) are respectively \(E\) closed unit ball and unit sphere.

We consider a normed vector space \(E\) over the field of the reals \(\mathbb{R}\) and a convex subset \(C \subset E\).

We suppose that \(0 \in C\) and that \(C\) is unbounded, i.e. there exists points in \(C\) at distance as big as we wish from \(0\).

The following question arises: “does \(C\) contains a ray?”. It turns out that the answer depends on the dimension of the space \(E\). If \(E\) is of finite dimension, then \(C\) always contains a ray, while if \(E\) is of infinite dimension \(C\) may not contain a ray. Continue reading An unbounded convex not containing a ray→